We need to prove that
$$\int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x > 1.$$
Fact 1: For all $y \in (0, \mathrm{e}]$,
$$\frac{-y^4 + 64y^3 - 318y^2 + 400y + 35}{132y + 48} \ge y(1 - \ln y)^2.$$
(Note: LHS is the (4, 1)-Pade approximant of $y(1 - \ln y)^2$ at $y = 1$.)
With the substitution $y = \mathrm{e}x$, using Fact 1, we have
\begin{align*}
& \int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x\\
=\, & \int_0^\mathrm{e} \frac{\mathrm{e}}{y^2 + \mathrm{e}^2 (1 - \ln y)^2}\, \mathrm{d} y\\
\ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2}\, \mathrm{d} y\\
\ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2]}\, \mathrm{d} y \tag{1} \\[8pt]
=\, &
\frac{1253273472\mathrm{e}}{911936989\mathrm{e}^2 - 120375682548} \int_0^\mathrm{e} \frac{1}{12y + 1}\, \mathrm{d} y\\[8pt]
&\quad + \frac{29129459463143424\mathrm{e}}{38952637956653\mathrm{e}^2 - 5141748210278196}
\int_0^\mathrm{e} \frac{1}{64y + 463}\, \mathrm{d} y \\[8pt]
&\quad + \frac{372594816\mathrm{e}}{2487336612548656836 - 18843459185974673\mathrm{e}^2}\\[8pt]
&\quad\quad \times \int_0^\mathrm{e} \frac{842186377401732y + 746090344086823}{1411344y^2 - 4426488y + 4934869}\, \mathrm{d} y \tag{2}\\
>\, & 1
\end{align*}
where in (1) we have used (easy): for all $y\in [0, \mathrm{e}]$,
\begin{align*}
&(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2\\
\le\, & (132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2].
\end{align*}
Note: The three integrals in (2) admit closed form expressions in terms of $\ln $ and $\arctan$ only.
We are done.